Problem: Let $R$ be the region enclosed by the polar curve $r(\theta)=2+\cos^2(6\theta)$ where $\dfrac{\pi}{6}\leq \theta\leq \dfrac{\pi}{3}$. $y$ $x$ $R$ $ 1$ $ 1$ Which integral represents the area of $R$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_0^{\scriptsize\dfrac{\pi}{3}}\left( 2+2\cos^2(6\theta)+\dfrac{1}{2}\cos^4(6\theta)\right)d\theta$ (Choice B) B $ \int_{\scriptsize\dfrac{\pi}{6}}^{\scriptsize\dfrac{\pi}{3}}\left( 2+2\cos^2(6\theta)+\dfrac{1}{2}\cos^4(6\theta)\right)d\theta$ (Choice C) C $ \int_0^{\scriptsize\dfrac{\pi}{3}}\left( 2+2\cos(6\theta)+\dfrac{1}{2}\cos^2(6\theta)\right)d\theta$ (Choice D) D $ \int_{\scriptsize\dfrac{\pi}{6}}^{\scriptsize\dfrac{\pi}{3}}\left( 2+2\cos(6\theta)+\dfrac{1}{2}\cos^2(6\theta)\right)d\theta$
Explanation: This is the formula for the area enclosed by a polar curve $r(\theta)$ between $\theta=\alpha$ and $\theta=\beta$ : $ \int_{\alpha}^{\beta}\dfrac{1}{2}\left(r(\theta)\right)^{2}d\theta$ Let's plug ${r(\theta)=2+\cos^2(6\theta)}$, ${\alpha=\dfrac{\pi}{6}}$, and ${\beta=\dfrac{\pi}{3}}$ into the formula and expand the parentheses: $\begin{aligned} &\phantom{=} \int_{\alpha}^{\beta}\dfrac{1}{2}\left({r(\theta)}\right)^{2}d\theta \\\\ &= \int_{{\scriptsize\dfrac{\pi}{6}}}^{{\scriptsize\dfrac{\pi}{3}}}\dfrac{1}{2}\left({2+\cos^2(6\theta)}\right)^{2}d\theta \\\\ &= \int_{\scriptsize\dfrac{\pi}{6}}^{\scriptsize\dfrac{\pi}{3}}\dfrac{1}{2}\left( 4+4\cos^2(6\theta)+\cos^4(6\theta)\right)d\theta \\\\ &= \int_{\scriptsize\dfrac{\pi}{6}}^{\scriptsize\dfrac{\pi}{3}}\left( 2+2\cos^2(6\theta)+\dfrac{1}{2}\cos^4(6\theta)\right)d\theta \end{aligned}$ In conclusion, this integral represents the area of region $R$ : $ \int_{\scriptsize\dfrac{\pi}{6}}^{\scriptsize\dfrac{\pi}{3}}\left( 2+2\cos^2(6\theta)+\dfrac{1}{2}\cos^4(6\theta)\right)d\theta$